Find Equivalence Classes for the relation $S$ on the set $\mathbb{R}^*$ is definied as $aSb \iff ab > 0$

Question

I am trying to understand how to determine the equivalence classes. I have this question in the book

The relation $S$ on the set $\mathbb{R}^*$ is definied as $aSb \iff ab > 0$

The answer in the book is $[1] = \mathbb{R}^+$ and $[-1] = \mathbb{R}^-$.

I don't even know where to start with this problem. I know that $ab > 0$ so then $a \cdot b$ always have to be positive and I know that equivalence class means disjoint categories but why are $[1]$ and $[-1]$ the equivalence classes?

Answer 1

First, for all $a,b \in \mathbb{R}^-$, $ab > 0$. So all the elements in $\mathbb{R}^-$ are in the same equivalence class.

Secondly, for all $a,b \in \mathbb{R}^+$, $ab > 0$. So all the elements in $\mathbb{R}^+$ are in the same equivalence class.

Finally, if $a \in \mathbb{R}^-$ and $b \in \mathbb{R}^+$, $ab < 0$. So $a$ and $b$ are not in the same equivalence class.

This shows you that the two equivalence classes (containing $\mathbb{R}^-$ and $\mathbb{R}^+$) are disjoint, so you have indeed two equivalent classes : $\mathbb{R}^-$ and $\mathbb{R}^+$.

Answer 2

Numbers $a$ and $b$ are equivalent with respect to relation $R$ if their product is positive. If you take $a>0$ what numbers are in relation with $a$? All $b>0$. Similarly for $a<0$, $b$ is in relation with $a$ if $ab>0$, so $b$ must be negative. $R$ partitions $\mathbb{R}^*$ into two nonempty, disjoint sets which add up to the whole $\mathbb{R}^*$, so the two equivalence classes are $\mathbb{R}_+$ and $\mathbb{R}_-$. Any positive number is in relation with $1$, so we can write $[1]_R=\mathbb{R}_+$, and similarly, we can denote $\mathbb{R}_-=[-1]_R$.